要求:输入两颗二叉树A和B,判断B是不是A的子结构
给出二叉树结构定义:
public class TreeNode<T> {
public T val;
public TreeNode<T> left;
public TreeNode<T> right;
public TreeNode(T val){
this.val = val;
this.left = null;
this.right = null;
}
分析:
由于B中有一部分和A是一样的,所以A是B的子结构,对于这个例子,我们可以分两步
- 第一步:在树B中找到和A的根节点值一样的节点
- 第二步:判断树B中以A根节点的子树是不是包含和树A一样的结构
首先B的根节点和A的根节点相同,接下来判断B和A的左右子树想不想同,很显然8!=9,那么我们可以放弃B的根节点,继续从B的左右子树两个方向判断,B的右子树淘汰7!=8,B的左节点符合要求,继续判断该节点的左右节点,符合要求,返回true
public class subTree {
public static boolean hasSubtree(TreeNode<Integer> root1, TreeNode<Integer> root2){
if(root2==null)
return true;
if(root1==null)
return false;
if(root1.val.equals(root2.val)){
if(tree1HasTree2FromRoot(root1,root2))
return true;
}
return hasSubtree(root1.left,root2) || hasSubtree(root1.right,root2);
}
public static boolean tree1HasTree2FromRoot(TreeNode<Integer> root1, TreeNode<Integer> root2){
if(root2==null)
return true;
if(root1==null)
return false;
if(root1.val.equals(root2.val) && tree1HasTree2FromRoot(root1.left,root2.left)
&& tree1HasTree2FromRoot(root1.right,root2.right))
return true;
else
return false;
}
public static void main(String[] args){
TreeNode<Integer> root1 = new TreeNode<>(8);
root1.left = new TreeNode<>(8);
root1.right = new TreeNode<>(7);
root1.left.left = new TreeNode<>(9);
root1.left.right = new TreeNode<>(2);
root1.left.right.left = new TreeNode<>(4);
root1.left.right.right = new TreeNode<>(7);
TreeNode<Integer> root2 = new TreeNode<>(8);
root2.left = new TreeNode<>(9);
root2.right = new TreeNode<>(2);
TreeNode<Integer> root3 = new TreeNode<>(2);
root3.left = new TreeNode<>(4);
root3.right = new TreeNode<>(3);
System.out.println(hasSubtree(root1,root2));
System.out.println(hasSubtree(root1,root3));
}
}
运行结果
true
false